All parts please 1) According
All parts please
1) According to a Field Poll, 79% of California adults (actualresults are 400 out of 506 surveyed) feel that “education and ourschools” is one of the top issues facing California. We wish toconstruct a 90% confidence interval for the true proportion ofCalifornia adults who feel that education and the schools is one ofthe top issues facing California.
a) The margin of error for a 90% confidence level is:_____(Write as a decimal rounded to the 4th decimal place)
b) A 90% confidence interval for the population proportion is:(_____ ,_____) (Write as a decimal rounded to the 3rd decimalplace.)
c) If we changed to an 85% confidence level, what would happento your confidence interval and your margin of error?
d) If we changed to a 95% confidence level, what would happen toyour confidence interval and your margin of error?
e) Five hundred eleven (511) homes in a certain southernCalifornia community are randomly surveyed to determine if theymeet minimal earthquake preparedness recommendations. One hundredseventy-three (173) of the homes surveyed met the minimumrecommendations for earthquake preparedness and 338 did not. Findthe confidence interval at the 90% Confidence Level for the truepopulation proportion of southern California community homesmeeting at least the minimum recommendations for earthquakepreparedness. (_____ , _____) (Round to the 4th decimal place)
f) State the previous confidence interval in the form of p^±ME._____ ± _____ (Round to the 5th decimal place)
g) Five hundred eleven (511) homes in a certain southernCalifornia community are randomly surveyed to determine if theymeet minimal earthquake preparedness recommendations. One hundredseventy-three (173) of the homes surveyed met the minimumrecommendations for earthquake preparedness and 338 did not. Thepoint estimate for the population proportion of homes that do notmeet the minimum recommendations for earthquake preparedness is_____ (Round to the 4th decimal place)
MULTIPLE CHOICE
What is meant by the term “90% confident” when constructing aconfidence interval for a proportion?
A.
If we took repeated samples, approximately 90% of the sampleswould produce the same confidence interval.
B.
If we took repeated samples, approximately 90% of the confidenceintervals calculated from those samples would contain the sampleproportion.
C.
If we took repeated samples, approximately 90% of the confidenceintervals calculated from those samples would contain the truevalue of the population proportion.
D.
If we took repeated samples, the sample proportion would equalthe population mean in approximately 90% of the samples.
Answer:
1) We are given that:
The sample standard deviation can be approximated to be equal to
The sampling distribution will have the same mean as thepopulation mean (which is unknown). Also, the standard deviation ofthe sampling distribution is given by:
A 90% confidence interval corresponds to a range of (0.05,0.95). Using a z-table, we get the corresponding z-score as (-1.65,1.65). So, we have:
So, the margin of error for a 90% confidence interval is0.0298.
b) So, the 90% confidence interaval for the population mean willbe:
So, a 90% confidence interval for the population proportion is(0.761, 0.820).
c) For 85% confidence interval, the corresponding range is(0.075, 0.925). Using a z-table, we get the z-score as (-1.44,1.44). So, we have margin of error 
. So, we have: 
. So, a 85% confidence interval for the population proportion is(0.764, 0.817). So, the margin oferror decreases and the confidence interval decreases when theconfidence level is decreased to 85%.
d) For 95% confidence interval, the corresponding range is(0.025, 0.975). Using a z-table, we get the z-score as (-1.96,1.96). So, we have margin of error 
. So, we have: 
. So, a 95% confidence interval for the population proportion is(0.755, 0.826). So, the margin oferror increases and the confidence interval increases when theconfidence level is increased to 95%.
e) We have:
Sample proportion 
Standard deviation 
Sample standard deviation 
For 90% confidence interval, the range is (0.05, 0.95). So, thez-score is (-1.65, 1.65).
So, the confidence interval will be 
f) As written above, we can also write it as:
g) The point estimate is directly calculated from the sample wehave gathered. So, the point estimate = 338/511 =0.6614.
h) What is meant by the term “90% confident” when constructing aconfidence interval for a proportion?
Answer: C. If we took repeated samples, approximately90% of the confidence intervals calculated from those samples wouldcontain the true value of the population proportion.
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